Question

Проиводные y=(sin3x)^x

Answer 1

$y=(\sin 3x)^x = e^{\ln (\sin 3x)^x} = e^{x \ln \sin 3x} \\ y'= e^{x \ln \sin 3x} \cdot (x \ln \sin 3x)' = (\sin 3x)^x \cdot ( \ln \sin 3x+\frac{3x \cos 3x}{\sin 3x} ) =\\ = (\sin 3x)^x \cdot ( \ln \sin 3x+3x ctg3x )$

Answer 2

Можно также применить логарифмическое дифференцирование:

Answer 3

To find the derivative of the function $y = (\sin(3x))^x$, we will use the chain rule. The chain rule states that if we have a function of the form $u = f(g(x))$, then its derivative is given by $du/dx = f'(g(x)) \cdot g'(x)$.

Let's break down the given function step by step to apply the chain rule:

1. Let $u = \sin(3x)$.
2. Let $y = u^x$.

Now, we need to find the derivatives of $u$ and $y$ with respect to $x$.

1. Derivative of $u = \sin(3x)$: Using the chain rule, the derivative of $u$ with respect to $x$ is: $\frac{du}{dx} = \frac{d}{dx}(\sin(3x)) = \cos(3x) \cdot \frac{d}{dx}(3x) = 3\cos(3x).$

2. Derivative of $y = u^x$: Now, we have $y = u^x$, and we need to find $dy/dx$. Using the chain rule again, we have: $\frac{dy}{dx} = \frac{d}{dx}(u^x) = \frac{du}{dx} \cdot x \cdot u^{x-1}.$

Now, substitute the value of $\frac{du}{dx} = 3\cos(3x)$ and $u = \sin(3x)$:

$\frac{dy}{dx} = 3\cos(3x) \cdot x \cdot (\sin(3x))^{x-1}.$

So, the derivative of $y = (\sin(3x))^x$ with respect to $x$ is $3\cos(3x) \cdot x \cdot (\sin(3x))^{x-1}$.